Change in Momentum and Impulse - AP Physics 1

A tennis ball experiences a constant force of magnitude $40\ \text{N}$ to the left for $0.020\ \text{s}$ . What is the magnitude of the impulse on the ball?
Answer: $0.80\ \text{N}\cdot\text{s}$ This problem tests understanding of change in momentum and impulse. Impulse is defined as the product of force and the time interval over which it acts: J = F·Δt. Given F = 40 N and Δt = 0.020 s, the impulse magnitude is J = (40 N)(0.020 s) = 0.80 N·s. The impulse-momentum theorem tells us that impulse equals the change in momentum of the object. Choice A incorrectly gives just the force value without considering time. To find impulse, always multiply the force magnitude by the time duration, ensuring your units are N·s.
A constant force of $7\,\text{N}$ downward acts on a toy for $0.30\,\text{s}$ . What is the toy's momentum change?
Answer: $2.1\,\text{N\!\cdot\!s}$ downward This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse measures the effect of force over time, J = F Δt. This is identical to the change in momentum, Δp = m Δv = J. The 7 N downward for 0.30 s leads to a momentum change of 2.1 N·s downward. Choice B provides the force without time, a typical oversight. Always multiply force by time interval to determine Δp effectively.
A $2.0\ \text{kg}$ glider receives an impulse of $-4.0\ \text{N}\cdot\text{s}$ (left). What is the glider's change in momentum $\Delta \vec p$ ?
Answer: $-4.0\ \text{kg}\cdot\text{m/s}$ (left) This problem tests understanding of change in momentum and impulse. The impulse-momentum theorem states that impulse equals change in momentum: J = Δp. Since the glider receives an impulse of -4.0 N·s (left), its change in momentum is exactly Δp = -4.0 kg·m/s (left). The units N·s and kg·m/s are equivalent for momentum. The negative sign indicates leftward direction. Choice B incorrectly divides the impulse by the mass, but impulse already equals the change in momentum directly. When given impulse, it directly equals the change in momentum without any additional calculations.
A $0.50\,\text{kg}$ cart initially moves right at $2.0\,\text{m/s}$ . A constant leftward force of $3.0\,\text{N}$ acts for $0.40\,\text{s}$ . What is the cart's change in momentum $\Delta \vec p$ ?
Answer: $-1.2\,\text{N}\!\cdot\!\text{s}$ (to the left) This problem tests understanding of change in momentum and impulse. The impulse-momentum theorem states that impulse (J = F·Δt) equals the change in momentum (Δp). Here, a leftward force of 3.0 N acts for 0.40 s, giving impulse J = (-3.0 N)(0.40 s) = -1.2 N·s (negative because force is leftward). Since impulse equals change in momentum, Δp = -1.2 N·s = -1.2 kg·m/s to the left. Choice A incorrectly uses positive direction, while choice D gives force instead of impulse. When calculating impulse, always multiply force by time duration, then recognize that impulse and change in momentum have identical values and units.
A $2.0\,\text{kg}$ cart is initially at rest. A constant rightward force of $6.0\,\text{N}$ acts for $0.50\,\text{s}$ . What is the cart's final momentum?
Answer: $3.0\,\text{kg}\!\cdot\!\text{m/s}$ to the right This problem requires applying the impulse-momentum theorem to find final momentum. The impulse delivered is J = F·Δt = (6.0 N)(0.50 s) = 3.0 N·s to the right. Since the cart starts from rest (initial momentum = 0), the change in momentum equals the final momentum: Δp = pf - pi = pf - 0 = 3.0 kg·m/s. Therefore, the final momentum is 3.0 kg·m/s to the right. Choice A incorrectly multiplies force by mass instead of time, while choice C gives force instead of momentum. Remember that impulse (N·s) and momentum (kg·m/s) have equivalent units, and for objects starting from rest, final momentum equals the impulse delivered.
A $0.50\,\text{kg}$ cart initially moves right at $2.0\,\text{m/s}$ . A constant $4.0\,\text{N}$ force acts left for $0.50\,\text{s}$ . What is $\Delta \vec p$ ?
Answer: $2.0\,\text{N\!\cdot\!s}$ to the left This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse is defined as the product of a constant force and the time interval over which it acts, given by J = F Δt. This impulse equals the change in momentum of the object, Δp = J, where momentum p = m v and the change accounts for direction as a vector. For the cart, the force of 4.0 N left for 0.50 s gives an impulse of 2.0 N·s left, which is the change in momentum regardless of initial velocity. A common distractor like choice A might result from incorrectly using initial momentum without considering direction. Always calculate impulse as F Δt and remember it directly gives Δp, including direction, to solve similar problems.
A $1.0\,\text{kg}$ cart receives a constant $15\,\text{N}$ push to the left for $0.10\,\text{s}$ . What is $\Delta \vec p$ ?
Answer: $1.5\,\text{kg\!\cdot\!m/s}$ left This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse arises from force acting over time, J = F Δt. It is synonymous with momentum change, Δp = J, incorporating direction. The 15 N left for 0.10 s produces Δp of 1.5 kg·m/s left. Choice D might stem from dividing instead of multiplying, an arithmetic slip. Verify by computing F Δt = Δp in every case for accuracy.
A $0.20\,\text{kg}$ ball's momentum changes by $+0.60\,\text{kg}\!\cdot\!\text{m/s}$ (rightward) during a collision lasting $0.030\,\text{s}$ . What is the average net force?
Answer: $+20\,\text{N}$ (rightward) This problem requires finding average force from change in momentum and time. The impulse-momentum theorem states J = Δp = F_avg·Δt, so we can solve for average force: F_avg = Δp/Δt. Given Δp = +0.60 kg·m/s and Δt = 0.030 s, we get F_avg = (0.60 kg·m/s)/(0.030 s) = 20 N rightward. The positive sign indicates rightward direction. Choice A incorrectly multiplies instead of dividing, while choice C divides by 1.0 instead of 0.030. To find average force during a collision, divide the momentum change by the collision duration.
A constant force of $3\,\text{N}$ acts downward on a glider for $2.0\,\text{s}$ . What is the magnitude of the glider's momentum change?
Answer: $6\,\text{kg\cdot m/s}$ This question assesses magnitude of momentum change from impulse in AP Physics 1. Impulse is constant force times time, representing the total momentum alteration. It equals Δp, with magnitude independent of direction for this query. The 3 N downward for 2.0 s gives a magnitude of 6 kg·m/s. Choice C is a distractor, possibly from dividing instead of multiplying. Apply J = F Δt directly for momentum change magnitudes in constant force situations.
A $2.0\,\text{kg}$ cart experiences a constant force $\vec F$ to the left for $0.30\,\text{s}$ , giving an impulse of $-1.8\,\text{N}\cdot\text{s}$ . What is $F$ ?
Answer: $-6.0\,\text{N}$ (left) This problem evaluates finding force from given impulse in AP Physics 1. Impulse equals force multiplied by time for constant forces, embodying the cumulative impact. It corresponds to change in momentum, so J⃗ = F⃗ Δt = Δp⃗. With J = -1.8 N·s over 0.30 s, F = -6.0 N to the left. Choice D is a distractor, perhaps from confusing impulse with momentum units. Rearrange the impulse formula to solve for unknowns like force in such scenarios.
A $2.0\,\text{kg}$ block initially at rest is pushed right by $10\,\text{N}$ for $0.30\,\text{s}$ . What is $\Delta \vec p$ ?
Answer: $3.0\,\text{kg\!\cdot\!m/s}$ right This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse is the force exerted multiplied by the time it acts, $J = F \Delta t$ . This quantity matches the change in momentum, $\Delta p = J$ , with direction aligning to the force. For the block at rest, the 10 N right for 0.30 s produces $\Delta p$ of $3.0 \, \text{kg}\cdot\text{m/s}$ right. Choice B errs by providing force alone, ignoring the time factor. Remember to multiply force by time to find impulse and equate it to $\Delta p$ as a general strategy.
A constant horizontal force of $10\,\text{N}$ acts on a cart for $0.10\,\text{s}$ . What time would be needed for a $5\,\text{N}$ force to produce the same impulse?
Answer: $0.20\,\text{s}$ This problem tests calculating time for equivalent impulses in AP Physics 1. Impulse is the product of constant force and time, reflecting the total effect on momentum. It equals the change in momentum, so same impulse means same Δp regardless of force-time combination. The 10 N force for 0.10 s gives 1 N·s, requiring 0.20 s for a 5 N force to match. Choice B is a distractor, possibly from halving time instead of doubling it. To solve similar problems, set impulses equal and solve for the unknown variable like time or force.
A constant net force of magnitude $8\,\text{N}$ acts on an object for $0.75\,\text{s}$ , in the direction of motion. What is the magnitude of $\Delta \vec p$ ?
Answer: $6\,\text{kg}\!\cdot\!\text{m/s}$ This question asks for the magnitude of momentum change given force and time. Using the impulse-momentum theorem, the magnitude of impulse is |J| = |F|·Δt = (8 N)(0.75 s) = 6 N·s. Since impulse equals change in momentum, |Δp| = 6 kg·m/s. The problem states the force acts in the direction of motion, so both impulse and momentum change are positive. Choice A incorrectly divides force by time, while choice D gives just the time value. When calculating momentum change from constant force, multiply force magnitude by time duration to get impulse magnitude.
A soccer ball experiences a constant $50\,\text{N}$ force to the right for $0.10\,\text{s}$ . What is the impulse $\vec J$ delivered to the ball?
Answer: $5.0\,\text{N}\!\cdot\!\text{s}$ to the right This question directly asks for impulse calculation. Impulse is defined as the product of force and the time interval over which it acts: J = F·Δt. With a rightward force of 50 N acting for 0.10 s, the impulse is J = (50 N)(0.10 s) = 5.0 N·s to the right. The impulse-momentum theorem tells us this impulse will equal the ball's change in momentum. Choice B incorrectly gives just the force value without multiplying by time. To find impulse, always multiply the force magnitude by the time duration and maintain the force's direction.
A tennis racket exerts an average force of $200\,\text{N}$ on a ball for $0.010\,\text{s}$ . What is the ball's momentum change magnitude?
Answer: $2.0\,\text{kg\cdot m/s}$ This question tests impulse and momentum change magnitude in AP Physics 1. Impulse is the average force times contact time, measuring momentum transfer. The theorem states this equals Δp, so magnitude is F Δt. The 200 N for 0.010 s gives a magnitude of 2.0 kg·m/s. Choice A is a distractor, likely from omitting time multiplication. For average force problems, multiply force by time to find momentum change reliably.
A $1.5\,\text{kg}$ block moving left receives a rightward impulse of $3.0\,\text{N}\cdot\text{s}$ . Which statement about $\Delta \vec p$ is correct?
Answer: $\Delta \vec p$ is $3.0\,\text{kg\cdot m/s}$ to the right This question examines the relationship between impulse and change in momentum in AP Physics 1. Impulse is force applied over time, serving as a vector quantity that changes an object's momentum. The theorem equates impulse directly to Δp⃗, meaning the change matches the impulse's magnitude and direction. Thus, a rightward impulse of 3.0 N·s causes Δp⃗ of 3.0 kg·m/s to the right, irrespective of initial motion. Choice A is a distractor as it wrongly assigns a leftward direction, perhaps confusing with initial velocity. Consistently use Δp⃗ = J⃗ to handle direction correctly in momentum problems.
A $0.25\ \text{kg}$ toy car experiences a constant force of $12\ \text{N}$ forward for $0.10\ \text{s}$ . What is the car's change in momentum?
Answer: $1.2\ \text{kg}\cdot\text{m/s}$ forward This problem tests understanding of change in momentum and impulse. Impulse equals force multiplied by time: J = F·Δt = (12 N)(0.10 s) = 1.2 N·s forward. The impulse-momentum theorem states that this impulse equals the change in momentum, so Δp = 1.2 kg·m/s forward. The mass of the car is not used in this calculation when force and time are given. Choice B incorrectly uses just the force value without considering time. When calculating change in momentum from a constant force, always multiply force by time duration to get impulse.
A hockey puck experiences a constant rightward force of $4\,\text{N}$ for $0.75\,\text{s}$ on frictionless ice. What is $\Delta \vec p$ ?
Answer: $3\,\text{kg\cdot m/s}$ to the right This problem evaluates the concept of change in momentum via impulse in AP Physics 1. Impulse represents the force exerted over a duration, quantifying how momentum is altered. According to the impulse-momentum theorem, the change in momentum $\Delta \vec{p}$ equals the impulse $\vec{J}$ , which is $\vec{F} \Delta t$ for constant force. On frictionless ice, the rightward 4 N force for 0.75 s produces $\Delta \vec{p}$ of $3 \, \text{kg}\cdot\text{m/s}$ to the right, regardless of initial velocity. Choice D is a distractor because it reverses the direction, possibly from misassigning signs. A key strategy is to align the direction of impulse with the applied force when calculating momentum changes.
A ball is initially at rest. A constant force of $6\,\text{N}$ acts upward on it for $0.20\,\text{s}$ . What is the magnitude of the impulse delivered?
Answer: $1.2\,\text{N\cdot s}$ This question assesses understanding of impulse and its relation to change in momentum in AP Physics 1. Impulse is the integral of force over time, but for a constant force, it simplifies to force multiplied by time. This impulse is equivalent to the change in momentum of the object, as per the impulse-momentum theorem. Here, the upward force of $6 \, \text{N}$ for $0.20 \, \text{s}$ results in an impulse magnitude of $1.2 \, \text{N}\cdot\text{s}$ , independent of the ball's mass since initial velocity is zero. Choice D is a distractor as it might result from multiplying force by time incorrectly or confusing units. Remember to compute impulse directly as $F \, \Delta t$ for constant forces to determine momentum changes accurately.
A $0.50\ \text{kg}$ cart initially moving right at $2.0\ \text{m/s}$ experiences a constant leftward force of $3.0\ \text{N}$ for $0.40\ \text{s}$ . What is the cart's change in momentum $\Delta \vec p$ ?
Answer: $-1.2\ \text{N}\cdot\text{s}$ (to the left) This problem tests understanding of change in momentum and impulse. Impulse is the product of force and time: J = F·Δt = (-3.0 N)(0.40 s) = -1.2 N·s, where negative indicates leftward direction. The impulse-momentum theorem states that impulse equals change in momentum: J = Δp. Therefore, Δp = -1.2 N·s (to the left). Choice C incorrectly multiplies force by initial velocity instead of time. When calculating impulse, always multiply force by the time duration, not by velocity or mass.

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